Question

Ionisation energy for hydrogen atom in the ground state is E. What is the ionisation energy of $L{i}^{++}$ atom in the $2^{nd}$ excited state

• A. E
• B. 3 E
• C. 6 E
• D. 9 E

Answer 1

The correct option is A E

Ionisation energy of atom in nth state $E_n\alpha \frac{Z^2}{n^2}$
For hydrogen atom in ground state (n = 1) and Z = 1 $\Rightarrow E=E_0$ ....... (i)
For $L{i}^{++}$ atom in $2^{nd}$ excited state n = 3 and Z = 3, hence $E^{{}^{\prime }}=\frac{E_0}{3^2}\times 3^2=E_0$ ........ (ii)
From equation (i) and (ii) $E^{{}^{\prime }}=E$